Is the affine hull of a set equal to the affine hull of the convex hull of that set

Question

Given a subset $S$ of $\mathbb{R}^n$, is it true that $\text{aff}(S) = \text{aff}(\text{co}(S))$? And if so how can I prove it rigidly?

Answer 1

Yes, it's true. Note first that $\operatorname{aff} \operatorname{co} S$ is an affine set containing $\operatorname{co} S$, and hence $S$ as well. By definition, $\operatorname{aff} S$ is the "smallest" affine set containing $S$, in that it is contained in any affine set containing $S$ (it can also be described as the intersection of all affine sets containing $S$). Thus, $$S \subseteq \operatorname{aff} S \subseteq \operatorname{aff} \operatorname{co} S.$$

To prove the other conclusion, note that every affine set is convex, which implies that $\operatorname{co} S \subseteq \operatorname{aff} S$. That is, $\operatorname{aff} S$ is an affine set containing $\operatorname{co} S$. Once again, the affine hull of $\operatorname{co} S$ is the smallest affine set that contains $\operatorname{co} S$, so it must lie between $\operatorname{co} S$ and $\operatorname{aff} S$, i.e. $$\operatorname{co} S \subseteq \operatorname{aff} \operatorname{co} S \subseteq \operatorname{aff} S.$$ This completes the proof.